physics I'M STUCK PZL HELP
posted by Andy .
A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influence of gravity. You can consider that the initial angular velocity is very small so that ω(θ=0∘)=0. The ruler has mass m= 100 g and length l= 35 cm. Use g=10 m/s2 for the gravitational acceleration, and the ruler has a uniform mass distribution. Note that there is no friction whatsoever in this problem. (See figure)
(a) What is the angular speed of the ruler ω when it is at an angle θ=30∘? (in radians/sec)
ω=
unanswered
(b) What is the force exerted by the wall on the ruler when it is at an angle θ=30∘? Express your answer as the x component Fx and the y component Fy (in Newton)
Fx=
unanswered
Fy=
unanswered
(c) At what angle θ0 will the falling ruler lose contact with the wall? (0≤θ0≤90∘; in degrees) [hint: the ruler loses contact with the wall when the force exerted by the wall on the ruler vanishes.]
θ0=

yes please.. that's the only question where I'm stuck too..

Me too!!!

Hi Teresa, can you do rocket problem me pls?

p: percentage (4.5% is burned in 155s)
m:mass of rocket
u:fuelspeed in meter/s!
v=u*ln(1/(1p))=1500*ln(1/0.955)
a=v/t = 69.065/155= 0.4455 
Hi Teresa, in my problem, question says The rocket burns 10 % of its mass in 290 s (assume the burn rate is constant).
What is the speed of the rocket after a burn time of 145.0 s?
How did you find 4.5% is burned in 155 sec? Pls.
Thank you very much 
because they ask you for half of the time,
so in your case in 145 it gets burned 5% in 145 seconds 
Thank you very much Teresa. I got green check. Now time for rular one..

in the equation above what is the "ln"?

"ln" is like log..it is in your calculator

is the "u" the ejected fuel in m/s instead of km/s?

hai Teresa can u help me for this Rocket question i don't know how to get 0.955 in the above ans !
Consider a rocket in space that ejects burned fuel at a speed of vex= 2.0 km/s with respect to the rocket. The rocket burns 11 % of its mass in 250 s (assume the burn rate is constant).
(a) What is the speed v of the rocket after a burn time of 125.0 s? (suppose that the rocket starts at rest; and enter your answer in m/s)
v=
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(b) What is the instantaneous acceleration a of the rocket at time 125.0 s after the start of the engines?(in m/s2)
a=